Lesson Overview
Chapter
12 — LPP
Days
6 Days · 9 Periods
Board Wt.
~10 marks ISC
Questions
20 + 14 MCQ
Sections Covered
Section A — Days 1–2
Linear Inequalities · Single & System · Half-planes · Feasible Region
Section B — Days 3–4
LPP Formulation · Manufacturing · Diet · Investment · Corner Point Method
Section C — Days 5–6
Graphical Solution · Maximise · Minimise · Bounded & Unbounded Regions
How to Use This Lesson
✏ Workspace
Graph questions split into left (calculation) and right (graph paper canvas) for drawing lines and shading regions.
💡 Step Solution
Manual: Click Next Step. Auto Play: Steps reveal at set intervals. ↺ Reset anytime.
⚡ Surprise Test
Teacher login required. Quick (3Q) / Day (5Q) / Chapter (10Q) / Custom. All 20 LPP questions included.
DAY 1 · SECTION A
Day 1 — Linear Inequalities: Single Inequality
📝ConceptDay 1 — Single Inequality▶ Show
1.1 Key Concepts — Linear Inequalities (Single)
HALF-PLANEA linear equation \(ax + by = c\) divides the plane into two half-planes. The inequality selects one.
SOLID LINEUse solid boundary for \(\leq\) or \(\geq\) — points on line ARE included in solution.
DASHED LINEUse dashed boundary for \(<\) or \(>\) — points on line are NOT in solution.
ORIGIN TESTSubstitute \((0,0)\) into the inequality. If true \(\rightarrow\) shade side containing origin. Else shade other side.
SHADINGThe shaded region represents all points satisfying the inequality.
Type 1 — Graph a Single Linear Inequality
Q 1Draw the graph of the solution set of \(3x - y \geq 1\) and indicate the solution region clearly.
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Type 2 — Strict vs Non-strict Inequalities
Q 2(i)Represent graphically: \(y > x\). Use a dashed line and shade the correct half-plane.
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Q 2(ii)Represent graphically: \(y \leq x\). Use a solid line and shade the correct half-plane.
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📚 Homework
Q 1: Also draw \(3x - y \leq 1\) and compare the two shadings. Q 2: Overlay both graphs on the same axes and describe the intersection region.📎 Assignments — Day 1 (click to expand)
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Day 1 — Video Resources
DAY 2 · SECTION A
Day 2 — System of Linear Inequalities
📝ConceptDay 2 — System of Inequalities▶ Show
2.1 Key Concepts — System of Linear Inequalities
SYSTEMA set of two or more inequalities that must ALL be satisfied simultaneously.
FEASIBLE REGIONThe common shaded area satisfying ALL inequalities is the feasible region.
CORNER POINTSPoints where boundary lines intersect. Found by solving pairs of boundary equations.
BOUNDEDFeasible region enclosed (finite area) \(\rightarrow\) bounded. Extends infinitely \(\rightarrow\) unbounded.
EMPTY REGIONIf no common region exists, the system has no solution.
Type 3 — System of Two Inequalities
Q 3Solve graphically: \(3x + 5y \leq 15\), \(x \geq 1\), \(x, y \geq 0\). Shade the feasible region and find all corner points.
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Type 4 — System of Three or More Inequalities
Q 4Solve graphically: \(y - 2 \geq 0\), \(x - y \geq 1\), \(5y - 2x \leq 10\), \(x, y \geq 0\). Shade feasible region and find all corner points.
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Type 5 — Read Constraints from a Given Graph
Q 5The shaded region in the given figure is the feasible region. Write the system of linear inequalities that produces it. [Upload figure using ✎ Edit Q → 🖼 Image]
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📚 Homework
Q 3: Verify all corner points by substituting back. Q 4: Which corner point gives the largest value of \(x + y\)?📎 Assignments — Day 2 (click to expand)
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DAY 3 · SECTION B
Day 3 — LPP Formulation I: Manufacturing & Diet
📝ConceptDay 3 — LPP Formulation I▶ Show
3.1 Key Concepts — LPP Formulation
DECISION VARIABLESLet \(x\) and \(y\) represent the quantities to be determined.
OBJECTIVE FUNCTION\(Z = ax + by\) is the linear function to be maximised (profit) or minimised (cost).
CONSTRAINTSLinear inequalities from resource limits: \(a_1 x + b_1 y \leq c_1\), etc.
NON-NEGATIVITY\(x \geq 0, y \geq 0\) always included — cannot produce negative quantities.
STEPSStep 1: Identify variables. Step 2: Form \(Z\). Step 3: List all constraints.
Type 6 — Manufacturing: Maximise Profit
Q 6A furniture dealer has ₹50,000 to invest and a store for at most 60 pieces. A table costs ₹2,500 and a chair ₹500. Profit: ₹250/table, ₹75/chair. Formulate an LPP to maximise profit.
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Q 7A company makes steel trunks A and B. Machine I: 1 trunk A or 2 trunks B per hour; available 18 h/week. Machine II: 3 trunks A or 1 trunk B per hour; available 12 h/week. Profit: ₹30 (A), ₹25 (B). Formulate the LPP.
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Type 7 — Diet Problem: Minimise Cost
Q 8A dietician mixes two foods to provide at least 8 units of vitamin A and 10 units of vitamin C. Food I: 2 units A & 1 unit C per kg (₹5/kg). Food II: 1 unit A & 2 units C per kg (₹7/kg). Formulate LPP to minimise cost.
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Q 9Fertilisers F1 and F2 contain 10% and 6% nitrogen, 6% and 10% phosphoric acid. Soil needs ≥14 kg nitrogen and ≥14 kg phosphoric acid. F1: ₹6/kg, F2: ₹5/kg. Formulate the LPP.
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📚 Homework
Q 6: Write the complete formulation with all constraints. Q 8: Identify the type of feasible region after formulating.📎 Assignments — Day 3 (click to expand)
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DAY 4 · SECTION B
Day 4 — Formulation II: Investment, Transport & Corner Point Method
📝ConceptDay 4 — Formulation II & Corner Points▶ Show
4.1 Key Concepts — Corner Point Method
CORNER POINT THEOREMIf optimal solution exists, it occurs at one (or more) corner point(s) of the feasible region.
FIND CORNERSSolve all pairs of boundary equations simultaneously. Check each lies in the feasible region.
EVALUATE ZCompute \(Z = ax + by\) at every corner point. Compare values.
MAX / MINThe largest Z-value is the maximum; smallest is minimum (bounded regions).
MULTIPLE OPTIMAIf Z equals at two adjacent corners, every point on that edge segment is also optimal.
Type 6B — Investment Problem
Q 10A person has ₹8,000 to invest in certificates A (6% p.a.) and B (8% p.a.). Minimum ₹500 in A, not more than ₹3,000 in B, total at least ₹2,000. Formulate LPP to maximise interest.
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Q 11A man has ₹1,500 to invest in stocks X (8%) and Y (10%). At least ₹200 in X, not more than ₹600 in Y. Formulate and solve graphically to maximise annual income.
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Type 7B — Transportation Problem
Q 12Plants P (300 bags/day) and Q (500 bags/day). Towns A, B, C need 250, 350, 200 bags. Transport costs (₹/bag): P→A=2, P→B=3, P→C=11, Q→A=1, Q→B=0, Q→C=6. Formulate LPP to minimise cost.
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Type 8 — Evaluate Objective Function at Given Corner Points
Q 13The corner points of the feasible region are \(O(0,0)\), \(A(5,0)\), \(B(4,3)\) and \(C(0,4)\). Find the maximum and minimum values of \(Z = 3x + 4y\).
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Q 14The corner points of a feasible region are \(A(0,10)\), \(B(5,6)\), \(C(15,0)\) and \(O(0,0)\). Find the minimum value of \(Z = 2x + y\).
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📚 Homework
Q 10, Q 12: Complete the formulation and solve graphically using Corner Point Method.📎 Assignments — Day 4 (click to expand)
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DAY 5 · SECTION B
Day 5 — Graphical Solution: Maximise
📝ConceptDay 5 — Graphical Solution: Maximise▶ Show
5.1 Key Concepts — Graphical Solution: Maximise
BOUNDED REGIONWhen feasible region is bounded (enclosed), both max and min of Z exist at corner points.
STEP 1Plot all boundary lines. Shade the region satisfying ALL constraints simultaneously.
STEP 2Find all corner points by solving boundary equations pairwise.
STEP 3Evaluate Z at each corner. The highest value is the maximum.
MULTIPLE OPT.If Z equals at two adjacent corners, all points on that edge are also optimal.
Type 9 — Graphical Solution: Maximise (Bounded Region)
Q 15Maximise \(Z = 2x + 3y\) subject to: \(x + 2y \leq 10\), \(2x + y \leq 14\), \(x \geq 0\), \(y \geq 0\). Find maximum Z and the point(s) where it occurs.
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Type 10 — Manufacturing: Full LPP (Maximise Profit)
Q 16A company makes FM sets (capacity 600/month) and AM/FM sets (capacity 525/month). One FM set: 30 man-hours; one AM/FM set: 40 man-hours; available: 24,000 man-hours/month. Profit: ₹180 (FM), ₹240 (AM/FM). Find: (i) max monthly profit; (ii) how many of each to produce.
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📚 Homework
Q 15: Solve again with \(Z = x + 2y\) and compare. Q 16: Verify corner point \((600, 150)\).📎 Assignments — Day 5 (click to expand)
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DAY 6 · SECTION B
Day 6 — Graphical Solution: Minimise & Unbounded Region
📝ConceptDay 6 — Minimise & Unbounded▶ Show
6.1 Key Concepts — Minimise & Unbounded Region
MINIMISEApply Corner Point Method. Evaluate Z at each corner. Smallest value is the minimum.
UNBOUNDEDWhen feasible region is unbounded, a minimum may or may not exist.
MIN TESTFind min value M at corner. Draw \(Z < M\) open half-plane. If NO common point with feasible region \(\rightarrow\) M is minimum. Otherwise, no minimum exists.
NO MAXIMUMFor maximise with unbounded growing region, there may be NO finite maximum.
SPECIAL CASECheck the direction of growing Z relative to the unbounded region.
Type 10B — Graphical Solution: Minimise
Q 17Minimise \(Z = x + y\) subject to: \(3x + 2y \geq 12\), \(x + 3y \geq 11\), \(x \geq 0\), \(y \geq 0\). Find minimum value and the point(s) where it occurs.
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Q 18Minimise \(Z = 20x + 10y\) subject to: \(x + 2y \leq 40\), \(3x + y \geq 30\), \(4x + 3y \geq 60\), \(x \geq 0\), \(y \geq 0\). Solve graphically.
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Type 11 — Unbounded Feasible Region
Q 19Maximise \(Z = -3x - 5y\) subject to: \(-2x + y \leq 4\), \(x + y \geq 3\), \(x - 2y \leq 2\), \(x \geq 0\), \(y \geq 0\). Comment on whether the maximum exists.
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Q 20Solve graphically: Maximise \(Z = x + 3y\) subject to: \(2x + 3y \geq 6\), \(x - y \geq 0\), \(x \geq 0\), \(y \geq 0\). Show that no maximum value of Z exists.
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📚 Homework
Q 17: Also find minimum of \(Z = 3x + 4y\) with the same constraints. Q 19: Is there a minimum? Justify.📎 Assignments — Day 6 (click to expand)
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📋 Multiple Choice Questions — Linear Programming
Section C — Multiple Choice Questions (1 mark each)
MCQ 1If the feasible region of an LPP is bounded, the objective function \(Z = ax + by\)
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Ahas only a maximum value
Bhas only a minimum value
Chas both maximum and minimum values
Dhas neither max nor min
MCQ 2The common region determined by all constraints of an LPP is called
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Aan unbounded region
Ban optimal region
Ca bounded region
Da feasible region
MCQ 3The point that lies in the half-plane \(2x + y - 4 \leq 0\) is
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A\((0, 8)\)
B\((1, 1)\)
C\((5, 5)\)
D\((2, 2)\)
MCQ 4Which point satisfies both \(2x + y \leq 10\) and \(x + 2y \geq 8\)?
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A\((-2, 4)\)
B\((3, 2)\)
C\((-5, 6)\)
D\((4, 2)\)
MCQ 5For \(Z = x + y\) subject to \(x + 2y \leq 70\), \(2x + y \leq 95\), \(x,y \geq 0\), the optimal solution is
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A\((20, 35)\)
B\((35, 20)\)
C\((30, 25)\)
D\((40, 15)\)
MCQ 6The objective function of an LPP is
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Aa constant
Ba linear function to be optimised
Can inequality
Da quadratic function
MCQ 7In the graphical method, the optimal solution of an LPP lies
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Aat the centre of the feasible region
Bat a corner point of the feasible region
Cat a point on the x-axis
Dat the origin
MCQ 8The maximum value of \(P = 5x + 4y\) subject to \(3x + 2y \leq 12\), \(2x + 3y \leq 13\), \(x,y \geq 0\) is
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A\(91/3\)
B\(31\)
C\(91\)
D\(20\)
MCQ 9Corner points: \((0,2), (3,0), (6,0), (6,8), (0,5)\). For \(Z = 4x + 6y\), minimum of Z occurs at
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A\((0,2)\) only
B\((3,0)\) only
Cmid-point of segment joining \((0,2)\) and \((3,0)\)
Dany point on segment joining \((0,2)\) and \((3,0)\)
MCQ 10\(Z = ax + by\) is maximum at \((8,2)\) and \((4,6)\). If \(a \geq 0, b \geq 0, ab = 25\), maximum value of Z is
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A\(60\)
B\(50\)
C\(40\)
D\(80\)
MCQ 11Corner points: \(O(0,0), A(250,0), B(200,50), C(0,175)\). If max of \(Z = 2ax + by\) occurs at both A and B, then relation between a and b is
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A\(2a = b\)
B\(2a = 3b\)
C\(a = b\)
D\(a = 2b\)
MCQ 12The solution set of \(3x + 5y < 4\) is
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Aopen half-plane not containing origin
Bopen half-plane containing origin
Cwhole XY-plane excluding the line
Dclosed half-plane containing origin
MCQ 13Corner points for \(x+y\leq 8\), \(2x+y\geq 8\), \(x,y\geq 0\): A\((0,8)\), B\((4,0)\), C\((8,0)\). If \(Z=ax+by\) has min on AB, then
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A\(8a+4=b\)
B\(a=2b\)
C\(b=2a\)
D\(8b+4=a\)
MCQ 14Feasible region corner points: \((0,0),(0,100),(25,100),(25,50)\). For \(Z=50x+15y\), maximum Z is
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A\(900\)
B\(1000\)
C\(1250\)
D\(2750\)